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STRUCTURE OF NIOBIUM ISOTOPES
By Prof. Lefteris Kaliambos (Natural Philosopher in New Energy) ( September 2014) Unfortunately the discovery of the assumed uncharged neutron (1932) along with the invalid relativity (EXPERIMENTS REJECT RELATIVITY) led to the abandonment of the well-established electromagnetic laws, in favour of various contradicting nuclear theories, which could not lead to the correct nuclear structure . Under this physics crisis and using the charged Up and Down quarks , discovered by Gell-Mann and Zweig , I published my paper “Nuclear structure is governed by the fundamental laws of electromagnetism ” (2003) by reviving the natural laws which led to my discovery of 288 quarks in nucleons including 9 charged quarks in proton and 12 ones in neutron able to give considerable charge distributions in nucleons for the discovery of nuclear force and structure by applying the laws of electromagnetism. (See my papers of nuclear structure in FUNDAMENTAL PHYSICS CONCEPTS ). Naturally occurring niobium (Nb), element 41, is composed of one stable isotope (Nb-93). Nb-93 is the lightest nuclide theoretically susceptible to spontaneous fission, and although this has never been observed, it makes niobium theoretically the lightest element with no stable isotope. The first 40 elements (to zirconium) all have at least one stable nuclide which is susceptible in theory only to proton decay. The most stable radioisotope is Nb-92 with a half-life of 34.7 million years. This nuclide is the longest-lived radionuclide of all elements that has not yet been detected in nature as a primordial isotope. (The nuclide with the next longest half-life, 244Pu with half-life 80 million years, has been detected, and is thus primordial). The next longest-lived niobium nuclides are Nb-94 (half-life: 20,300 years), and Nb-91 with a half-life of 680 years. There is also a meta state at 31 keV whose half-life is 16.13 years. Twenty three other radioisotopes have been characterized. Most of these have half-lives that are less than two hours except Nb-95 (35 days), Nb-96 (23.4 hours) and Nb-90 (14.6 hours). ' ' STRUCTURE OF Nb-83, Nb-101, Nb-103, Nb-105, Nb-107, Nb-109, Nb-111, AND Nb-113 WITH S = +5/2 ' For understanding the structure of the above unstable nuclides with an odd number of extra neutrons we see that they are based on the structure of Nb-83 with S = + 5/2 which changes the structure of the following diagram of Nb-82 when the extra n42(+1/2) fills the blank position formed by p38 and p31. ' ''' '''DIAGRAM OF Nb-82 WITH S =0 This structure is based on the parallelepiped of Mg-24 having six horizontal planes of opposite spins like the +HP1, -HP2, +HP3, -HP4, +HP5, and -HP6. We also see the horizontal squares of negative spins and positive ones like the -HSQ and the +HSQ . Here the deuterons existing from p13n13 to p20n20 and the deuterons existing from p33n33 to p36n36 are not shown because they are in front of the Mg-24 or behind it. Note that all the 40 deuterons existing from p1n1 to n40p40 give a total spin S= 0 and in the structure of Nb-82 the additional p41(-1/2) fills the blank position formed by n39 and n21, while the additional n41(+1/2) fills the blank position formed by p40 and p32. ' ' ' n40.......p40.....n41' ' +HSQ p38.......n38 ' ' n31………p12........n12......p32' ' -HP6 p31....n11.........p11…… n32 ' ' p29....... n10.........p10……n30' ' +HP5 n29……p9..........n9 …….p30 ' ' n27.........p8..........n8.......p28'' ' -HP4 p27.....n7..........p7.......n28 ' ' p25.........n6.........p6.......n26' ' +HP3 n25……p5........n5……...p26 ' ' n23………p4........n4……..p24' ' -HP2 p23…….n3…….p3……….n24 ' ' p21.........n2………p2........n22' ' +HP1 n21......p1........n1........p22 ' ' p37......n37 ' ' -HSQ p41....n39......p39 ' Now under the one extra n42(+1/2) which gives the structure of Nb-83 we see that the n37p37 changes the spin from S =-1 to S =+1 because it goes from the -HSQ to +HSQ to make horizontal bonds in front of p38n38. Note that this change of spins gives S = +2 . Since the one extra n42(+1/2) fills the blank position between the p38 and p31, the total spin of Nb-83 is given by S = +2 +1(+1/2) = +5/2 Under this structure of Nb-83 with S =+5/2 we see that the structure of the above unstable nuclides with an odd number of extra neutrons is based on the Nb-83 with +5/2 . It gives a large number of extra neutrons which make single bonds leading to the decay . For example the Nb-101 with S= +5/2 has 18 more extra neutrons of opposite spins. ' ' STRUCTURE OF Nb-85, Nb-89, Nb-91, Nb-93, Nb-95, Nb-97 AND Nb-99 For understanding the structure of the above nuclides with the same odd number of extra neutrons you must read my STRUCTURE OF Nb-93 with S = +9/2. After a careful analysis we found that the nuclides of this group are based on a new structure of Nb-85, which has not the well-known structure of S = +5/2 but another structure of S = +9/2. In these structures with an odd number of extra neutrons of the above nuclides the protons make blank positions for receiving extra neutrons which make two bonds per neutron but the small number of neutrons in the unstable Nb-85, Nb-89, and Nb-91 cannot give enough binding energies to pn bonds for overcoming the pp and nn repulsions. However in the stable Nb-93 the greater number of neutrons gives enough binding energies to pn bonds which lead to the stability. Whereas the two more neutrons in the unstable Nb-95 under the absence of blank positions make single bonds leading to the decay. Similarly, the extra neutrons than those of Nb-93 in the unstable Nb-97 and Nb-99 make the same single bonds leading to the decay. ' ' STRUCTURE OF Nb-82, Nb-87, Nb-98, Nb-100, Nb-102, AND Nb-104 ' Under the structure of Nb-82 with S = 0, the Nb-87 with S =-1/2 has 3 extra neutrons of negative spins and 2 extra neutrons of positive spins . That is S = 0 +3(-1/2) + 2(+1/2) = -1/2 Whereas the Nb-98, Nb-100, Nb-102, and Nb-104 with S =+1 have 2 extra neutrons of positive spins and an even number of extra neutrons of opposite spins giving S =0. For example the Nb-104 has 2 extra neutrons of positive spins and 20 extra neutrons of opposite spins giving S =0. That is S = 0 + 2(+1/2) + 0 = +1. ' ''' '''STRUCTURE OF Nb-84, Nb-106, Nb-108, Nb-110, AND Nb-112 Using again the diagram of Nb-82 we see that in the presence of an even number of extra neutrons the p37n37 with S = -1 goes to p38n38 for making horizontal bonds with S = +1. So this change of spins gives S= +2. Also the p41 (-1/2) goes to n41(+1/2) giving S = +1. Under this new structure of Nb-82 with S=+3 the two extra neutrons of opposite spins in the structure of Nb-84 give S =+3. Therefore in the presence of extra neutrons more than those of Nb-84 we see that the structure of the above nuclides is based on the structure Nb-84 with S = +3 . For example the Nb-108 with S = +2 has two more extra neutrons of negative spins and 22 extra neutrons of opposite spins giving S =0. That is the total spin of Nb-108 is given by S = +3 + 2(-1/2) + 0 = +2 ' ' STRUCTURE OF Nb-86, Nb-88, Nb-90, Nb-92, Nb-94, AND Nb-96 In the presence of extra neutrons as those of the above unstable nuclides the deuterons p37n37 and p39n39 change their spins from S = -2 to S = +2 giving S = +4 because they go from the –HSQ to +HSQ in order to make horizontal bonds with p38n38 and p40n40. Also at the same +HSQ the additional p41n41 with S = +1 makes horizontal bonds with n38p40. In other words we have a new structure of Nb-82 with S = +5. Under this condition in the presence of extra neutrons the structure of the above unstable nuclides is based on the new structure of Nb-82 with S = +5 . For example the Nb-86 with S = +6 has 2 extra neutrons with positive spins and 2 extra neutrons of opposite spins giving S =0. That is S = +5 + 2(+1/2) + 0 = +6 STRUTURE OF Nb-81 WITH S =-3/2 In the absence of one neutron of positive spin we see that the structure of Nb-82 has S = -1. In this case the additional p41n41 with S = -1 makes horizontal bonds with the nucleons of the -HSQ. Thus in the structure of Nb-81 with S = -3/2 the spin is given by S = -1 - (+1/2) = -3/2 . Category:Fundamental physics concepts